# Battery life calculation.

Discussion in 'Power Electronics' started by cps13, Sep 29, 2017.

1. ### cps13

54
1
Feb 25, 2013
Hi,

Please can someone help me with calculating how much battery capacity I need. Calculating different ways which all seem logical in some respects, I can get very different answers! What I have is:

- A datalogger which takes a sample 3 times a minute.
- The sample time is ~1sec and draws 2mA.
- When it is not sampling it draws 2uA.
- I need the battery to last for a year. I have done:

1 second x 3 = samples per minute at 2mA. = 3
3 x 60 = 180
180 x 24 = 4320 (amount of 2mA current draw per day in seconds)
4320 x 365 = 1576800 seconds

1576800 / 60 = 26280 minutes at 2mA per year
26280 / 60 = 438 hours at 2mA per year

438 hours x 0.002 = 0.876Ah

8760 hours per year - 438 = 8322 hours at rest current
8322 x 0.000006 = 0.016644 Ah requirement

Total requirement = 0.016644 + 0.876 = 0.892644 Ah.

Please can anyone confirm if I have done this correctly?

Thanks,

2. ### kellys_eye

4,713
1,308
Jun 25, 2010
Many other variables to consider too - have you checked the battery chemistries and/or consulted the manufacturers data sheets for the proposed battery?

Your calculations are based on 100% efficiency, perfect ambient temperature, no self discharge, an 'ideal' battery etc etc. (Rounded up to 1Ahr also makes the maths easier..... the calculations are otherwise ok as far as my efforts are concerned!)

But allowing for inefficiencies etc etc I reckon you'd need 'multiples' of that capacity and/or maybe 'joule-thief' technology to ensure full battery usage.

3. ### Harald KappModeratorModerator

12,447
2,981
Nov 17, 2011
I do not follow your calculation, it seems arther too complex for this purpose. However, your result is in the correct range.

This is how I'd calculate:
The average current within one minute (incl. idle current) is: ((2 mA * 3 s) + (0.002 mA * 57 s)) / 60 s = 0.102 mA (rounded)
1 year has 24*356 hours, therefore the energy used within 1 year is E(1y) = 8544 h * 0.102 mA = 871 mAh (rounded)
Add 15 % reserve (typically you cannot get 100 % of the rated capacity from a battery), then you arrive at 1 Ah.

4. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
2,852
Jan 21, 2010
Over a year you might also need to consider self-discharge.

Another issue that falls in your favour is that most batteries have measured capacity based on a 20 hour discharge. You will be discharging at a much lower rate and that will result in a larger useable capacity.

5. ### cps13

54
1
Feb 25, 2013
Thanks - yep I will factor in more capacity for temperature effects etc. but I just wanted to check my initial calculation was correct.

thanks

6. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,505
2,852
Jan 21, 2010
It seems right.

Everyone, it seems has their preferred approach in calculating this.

I would calculate average current in amps and multiply it by the number of hours.

Average current = (3/60 * 2e-3) + (57/60 * 2e-6) = 1.019e-4 amps

Hours = 24*365 = 8760

So Ah = 1.019e-4 * 8760 = 0.8926

(@Harald Kapp used 356 days in a year - we have longer years here)

My method is the same as Harald's other than I calculated in amps vs milliamps, and the length of the year . Somehow I read his method as different.

7. ### Harald KappModeratorModerator

12,447
2,981
Nov 17, 2011
umpfh
We do, too

Arouse1973 likes this.